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zhannawk [14.2K]
2 years ago
7

The equation (x + 6)^2 + (y + 4)^2 = 36 models the position and range of the source of a radio signal. Describe the position of

the source and the range of the signals
Mathematics
1 answer:
Ilia_Sergeevich [38]2 years ago
7 0

Answer:

position: (-6, -4)

range: 6

Step-by-step explanation:

The equation is that of a circle centered at (-6, -4) with a radius of √36 = 6. We presume that the "position" is that of the circle's center, and the "range" is the radius of the circle.

___

The standard form equation of a circle with center (h, k) and radius r is ...

(x -h)^2 +(y -k)^2 = r^2

Matching parts of the equation, we find ...

h = -6, k = -4, r = √36 = 6.

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Help me with these math questions....
Nady [450]

Answer: cotθ

<u>Step-by-step explanation:</u>

 tanθ * cos²θ * csc²θ

=  \dfrac{sin\theta}{cos\theta} * \dfrac{cos\theta*cos\theta}{} *\dfrac{1}{sin\theta*sin\theta}

= \dfrac{cos\theta}{sin\theta}

= cotθ

Answer: B

<u>Step-by-step explanation:</u>

The parent graph is y = x²

The new graph y = -x² + 3 should have the following:

  • reflection over the x-axis
  • vertical shift up 3 units

Answers:

  • a. Quadrant II
  • b. negative
  • c. \dfrac{\pi}{6}
  • d. C
  • e.-\dfrac{\sqrt{3}}{3}

<u>Explanation:</u>

\dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}

a) Quadrant 2 is: \dfrac{\pi}{2} < \theta < \pi

b) In Quadrant 2, cos is negative and sin is positive, so tan is negative

c) \pi-\dfrac{5\pi}{6} = \dfrac{\pi}{6}

d) the reference line is above the x-axis so it is negative -->  -tan\dfrac{\pi}{6}

e) tan(\dfrac{5\pi}{6})=\dfrac{1}{-\sqrt{3}}=-\dfrac{\sqrt{3}}{3}


4 0
2 years ago
Solve for B -(12 x 2) x (-6) = -12 x (2 x b)<br> A. -6<br> B. 6<br> C. -12<br> D. 12
myrzilka [38]

<u><em>Answer:</em></u>

<u><em>Option A</em></u>

<u><em>Step-by-step explanation:</em></u>

<u><em>-(12 x 2) x (-6) = -12 x (2 x b)</em></u>

<u><em>=> -24 x -6 = -12 x 2b</em></u>

<u><em>=> 144 = -24b</em></u>

<u><em>=> b = -6</em></u>

<u><em>Hoped this helped.</em></u>

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